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On 12/5/2024 2:30 PM, WM wrote:Your following exposition is lucid and clear. It was a pleasure for me to discuss it!
Yes.>Depending upon how 'end.segment' is defined,
And it is the empty endsegment.
{} either is or isn't an end.segment.
Consider the options.
⎛ With {} as an end.segment,
⎜ there are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎜ the intersection of all holds no finite cardinal.
⎜all endsegments hold content.
⎜ With {} NOT as an end.segment,
⎜ there STILL are more.than.finite.many end.segments,Not actually infinitely many however. If all endsegments have content, then not all natnumbers are indices, then the indices have an upper bound.
⎜ too many for any finite.cardinal to beToo many for all definable natnumbers. But by assumption of content not all natnumbers have become indices.
⎜ in common with all end segments.
⎜ And therefore,No definable finite cardinal.
⎝ the intersection of all STILL holds no finite cardinal.
The intersection of all non.empty.end.segmentsThat is a wrong conclusion because inclusion monotony prevents an empty intersection of non-empty endsegments:
of the finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
Because,That is true only for definable or accessible cardinals.
⎛ for each finite.cardinal,
⎜ there are fewer finite.cardinals before it
⎝ than there are finite.cardinals after it.
Not becauseThis argument is wrong if infinite bijections are assumed to exist. However "fewer finite.cardinals before it than there are finite.cardinals after" is the your only argument.
( an end.segment is empty.
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